Surreal numbers

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How do you show that the surreal numbers form a proper class?
 

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  • #2
matt grime
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I suspect it goes along the lines of: show that there is proper class of ordinals, and show that for each ordinal one can construct inequivalent surreal numbers, or something.

Since there are 'more' ordinals than cardinals, and there is a proper class of cardinals (if there is a set of cardinals, C, what is the cardinality of the power set of C?), this becomes easy - if I've understood the one thing I read about surreal numbers and some link to ordinals, and that;s a big if.
 
  • #3
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I don't understand why the surreal numbers don't form a complete ordered field, in which case the real numbers wouldn't be the only complete ordered field. I guess technically if the surreals form a proper class then it wouldn't count, but that's... wrong, somehow.
 
  • #4
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Never mind, I was reading the Wiki article, and I apparently misunderstood some things.
 
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  • #5
matt grime
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I don't know what you've read so I can't really comment, but completeness usually requires some notion of a metric - i.e. a mapping into R, the real numbers. So what is the norm of any non-real surreal?
 
  • #6
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If a metric needs to be real-valued, and if the surreals form a proper class (or indeed a set bigger than reals) then it can't be defined.

But I haven't read anything on the "norm" of surreal numbers; maybe it's defined differently.
 
  • #7
matt grime
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If a metric needs to be real-valued,
It does.


and if the surreals form a proper class

they do - apparently.

Checking the definitions is always a good idea.
 
  • #8
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I guess that answers the question about complete ordered fields.
 
  • #9
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(Rehash of my earlier post... only this time, correct)
Ignoring that the surreals form a proper class rather than a set, you'd need to prove that you can't define a metric on them that turns them into a complete ordered field.

One way to see that the surreal numbers are not a complete ordered field is to use the axioms of a complete ordered field to show that they satisfy the archimedean property, which says that there are no infinitesimals other than 0 and no infinities. Since the surreal numbers have non-zero infinitesimals and infinities, they cannot be a complete ordered field.
 
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  • #10
Hurkyl
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The easiest way to show the surreals are a proper class is by contradiction. Assume they form a set, then construct a new surreal number not in that set.

(hint... can you find one bigger than everything in that set?)
 
  • #11
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If X is the set of all surreal numbers, then so is [tex]\{ X|\emptyset\}[/tex]. But this only leads to a contradiction assuming well-foundedness.
 
  • #12
Hurkyl
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If X is the set of all surreal numbers, then so is [tex]\{ X|\emptyset\}[/tex]. But this only leads to a contradiction assuming well-foundedness.
Or by trichotomy....
 
  • #13
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How does it violate trichotomy?
 
  • #14
Hurkyl
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You've asserted:

X is the set of all surreal numbers
{X | } is a surreal number

and it's a basic fact of surreal numbers that

{X | } is larger than every member of X

so....


Now that I think of it, I'm pretty sure it's also a theorem that every surreal number is a well-founded set.
 
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